# Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions. y''-7y'+6y =e^t + delta(t-3) + delta(t-5) y(0) = 0 y'(0) = 0

Question
Laplace transform
Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions.
$$y''-7y'+6y =e^t + \delta(t-3) + \delta(t-5)$$
$$y(0) = 0$$
$$y'(0) = 0$$

2021-01-20
Step 1
Here we use table of Laplace transforms and some properties of Laplace transform.
Step 2
$$y''-7y'+6y =e^t + \delta(t-3) + \delta(t-5)$$
$$y(0) = 0$$
$$y'(0) = 0$$
Apply laplace transform $$\Rightarrow L\left\{y"\right\}-7L\left\{y'\right\}+6L\left\{y\right\}=L\left\{e^t\right\}+L\left\{\delta(t-3)\right\}+L\left\{\delta(t-5)\right\}$$
$$\text{use } L\left\{\delta(t-c)\right\}=e^{-cs}$$
$$\Rightarrow (s^2Y(s)-sY(0)-y'(0))-7(sY(s)-y(0))+6y(s)=\frac{1}{(s-1)}+e^{-3s}+e^{-5s}$$
$$\Rightarrow (s^2-7s+6)Y(s)=\frac{1}{(s-1)}+e^{-3s}+e^{-5s}$$
$$\Rightarrow Y(s)=\frac{1}{(s-1)(s^2-7s+6)}+\frac{e^{-3s}}{(s^2-7s+6)}+\frac{e^{-5s}}{(s^2-7s+6)}$$
Now apply inverse laplace transform
$$\Rightarrow L^{-1}\left\{Y(s)\right\}=L^{-1}\left\{\frac{1}{(s-1)(s^2-7s+6)}\right\}+L^{-1}\left\{\frac{e^{-3s}}{(s^2-7s+6)}\right\}+L^{-1}\left\{\frac{e^{-5s}}{(s^2-7s+6)}\right\}$$
Now use paricl and use table of Laplace transforms
$$\Rightarrow y(t)=-\frac{1}{6}t \cdot e^t - \frac{1}{6} e^(t-3)\cdot \delta(t-3)-\frac{1}{6}e^{t-5}\cdot \delta(t-5)$$
here we use properties $$L^{-1}\left\{F(s-c)\right\}=e^{ct} \cdot f(t)$$
$$L^{-1}\left\{e^{-cs} \cdot F(s)\right\}=y(t) \cdot f(t)$$

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